a) Find all the values of $\alpha$ such that $f'(0)$ exists. b) Find all the values of $\alpha$ such that $f$ is of bounded variation on $[0,1]$ For any positive real numbers $ \alpha $ and $\beta$, define 
$f(x) = \begin{cases} x^{\alpha} \sin\frac{1}{x^\beta} && \text{if $x \in (0,1]$,}\\
           0 && \text{if $x = 0$}\end {cases}$
a) For a  given $\beta  > 0$ , find all the values of $\alpha$  such that $f'(0)$ exists.
b) For  given $\beta >0,$ find all the values of $ \alpha$  such that $f$ is  of bounded variation on $[0,1]$
My ANSWER: For $ a) $ $\alpha \ge \beta$  then  $f$  is bounded variation .  now  $ f'(0) $will exists  if  $\alpha \ge \beta \ge 0$
For $b)$ same condition as  for $(a)$
Edit answer :$f$ has derivative $$\displaystyle f^\prime(x) = \begin{cases} \alpha x^{\alpha-1} \sin\left(\dfrac{1}{x^\beta}\right) - \dfrac{x^\alpha}{x^{\beta +1}} \cos\left(\dfrac{1}{x^\beta}\right) &\text{on }(0,1], \\\\ 0 & \text{if }x = 0.
\end{cases}$$ Hence $$\vert f^\prime(x) \vert \le \alpha x^{\alpha-1} + x^{\alpha - \beta -1}$$ The integrals $\int_0^1 x^{\alpha-1}dx$ and $\int_0^1 x^{\alpha - \beta -1}dx$ both converge for $1 < \alpha < 1 +\beta $. Hence $$V_0^1(f) \le \int_0^1 \vert f^\prime(x) \vert dx$$ and $f$ is of bounded variation on $[0,1]$ as the RHS of the inequality is finite.
Is  my  answer correct ??? or incorrect ??  Please  rectify it.
 A: For the first part $f'(0)$ exists if $\displaystyle\lim_{h\rightarrow 0^+}\dfrac{f(0+h)-f(0)}{h}$ exists. Without loss of generality fix $\beta>0$, we have the limit $$\displaystyle\lim_{h\rightarrow 0^+}\dfrac{(0+h)^{\alpha}\sin\Big(\dfrac{1}{(0+h)^{\beta}}\Big)-0}{h}=\lim_{h\rightarrow 0^+}h^{\alpha-1}\sin\Big(\dfrac{1}{h^{\beta}}\Big) .$$ Using squeeze theorem we see the limit exists if $\alpha\geq \beta+1$.
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For the second part we calcualte $f'(x)$.
$$ f'(x)=\begin{align}\begin{cases}\alpha x^{\alpha-1}\sin\Big(\dfrac{1}{x^{\beta}}\Big)-\dfrac{\beta x^{\alpha}}{x^{\beta+1}}\cos\Big(\dfrac{1}{x^{\beta}}\Big), 
 &x\neq0\\0, &x=0\end{cases}\end{align}$$ If total variation is finite we say $f$ is of bounded variation. Also if $f$ is differentiable and its derivative is Riemann-integrable, its total variation is given by 
$${\displaystyle V_{a}^{b}(f)=\int _{a}^{b}|f'(x)|\,\mathrm {d} x.} .$$
$$V_0^1(f)=\int_0^1|f'(x)|dx\leq\alpha\int_0^1x^{\alpha-1}dx+\beta\int_0^1x^{\alpha-\beta-1}dx=1+\dfrac{\alpha}{\alpha-\beta} $$
(The above integrals Riemann integrable only if  $\alpha-1>-1$ and $\alpha-\beta-1>-1$). Thus the total variation $V_0^1(f)$ is finite if $\alpha>0$ and $\alpha>\beta$.
A: As explained in Yadati Kiran's answer, if $\alpha > \beta > 0$, then $f'$ is integrable. By the Fundamental Theorem of Calculus, we have
$$
\int_y^x f' = f(x) - f(y)
$$
for any $x > y > 0$.
Taking the limit as $y \to 0$, we obtain
$$
f(x) = \int_0^x f'
$$
by continuity of integration.
Now let $P = \{x_0, \cdots, x_k\}$ be a partition of $[0,1]$.
We then have
\begin{align*}
V(f, P) &= \sum_{i=1}^k |f(x_i) - f(x_{i-1})| \\
 &= \sum_{i=1}^k \left|\int_0^{x_i} f' - \int_0^{x_{i-1}} f'\right| \\
&= \sum_{i=1}^k \left|\int_{x_i}^{x_{i-1}}f'\right| \\&\leq \sum_{i=1}^k \int_{x_{i-1}}^{x_i}|f'| \\
&= \int_0^1 |f'| \\
&< \infty
\end{align*}
Since $P$ was arbitrary, $f$ is of bounded variation.
Now suppose $\beta \geq \alpha > 0$. 
Fix an index $n$ and consider the following partition:
$$
P_n = \left\{0, \left(\frac{2}{2 n \pi}\right)^{1 / \beta}, \left(\frac{2}{(2n - 1) \pi}\right)^{1 / \beta}, \cdots, \left(\frac{2}{\pi}\right)^{1 / \beta}, 1 \right\}
$$
It can be shown
\begin{align*}
V(f, P_n) &= c \sum_{k=0}^{n - 1} \frac{1}{(1 +
2k)^{\alpha / \beta}} + \sin(1) - c \\
&\geq \frac{c}{2^{\alpha/\beta}} \sum_{k=1}^{n} \frac{1}{k^{\alpha / \beta}} + \sin(1) - c
\end{align*}
where
$$
c = 2 \left(\frac{2}{\pi}\right)^{\alpha / \beta}
$$
Since $0 < \alpha / \beta \leq 1$, the series on the right-hand side diverges. 
Therefore $f$ is not of bounded variation. 
